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Offline Mr Andrew

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The momentum operator
« on: 07/06/2008 19:08:33 »
In quantum mechanics, all observable quantities are defined by eigenvalues of Hermitian operators.  For momentum the operator is -i*(h/2π)*∂/∂x.  Why is that the operator and not simply p0?  For position the operator is simply x...why can't it be the same way for momentum?


 

Offline JP

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The momentum operator
« Reply #1 on: 07/06/2008 19:29:49 »
The explanation depends on how deep you want to go into the mathematics of it.  The simplest explanation is that the uncertainty principle means you can't get a single number for p if you have a single number for x.  So if your operator for position is just "x," your operator for momentum has to be messier.  I think the better explanation is to dig just a little deeper and understand that your wavefunction could either be as a function of position or as a function of momentum: you can write it as either Ψ(x) or Ψ(p), and the form your operators take depend on the choice of this variable.

If you define your wavefunctions as functions of position, then the operators you use are:
x=x
p=-i*(h/2π)*∂/∂x
Ψ=Ψ(x)
Boldface here is being used for the general operators and wavefunction, and non-boldface is being used for operators once you choose to work as functions of position or functions of momentum.

But you can also think of them as functions of momentum, in which case:
x=i*(h/2π)*∂/∂p
p=p
Ψ=Ψ(p)

It turns out these are the relations that satisfy the uncertainty relation.  If you're more mathematically savvy, they're all tied up in Fourier transforms, which is how you change from variables x to p.
 

Offline Mr Andrew

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The momentum operator
« Reply #2 on: 08/06/2008 03:26:38 »
Ah, Fourier Transforms!  I forget how those work.  I could probably go on wikipedia and find out but most of the time the math pages are not so user-friendly there.
 

Offline JP

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The momentum operator
« Reply #3 on: 08/06/2008 05:36:03 »
The math behind Fourier transforms tends to get fairly non-user-friendly.  But the basic idea is that you can express some physical process as a function or one variable, or second variable.  You can change between these by Fourier transformations.  In this particular case, your wavefunction Ψ(x) tells you the probability of measuring a particle at different positions, x.  If you want to know the probability of measuring a particle with specific momentums, p, you need to Fourier transform and look at the wavefunction Ψ(p). 

This also shows up in other areas of physics.  For example, you can express an (electromagnetic) signal as a waveform in time Ψ(t), or Fourier transform it and find out how much amplitude was given to each Frequency present in the wave, Ψ(ν), where ν is the frequency.

If you want to try to figure out the details, they're given here: http://mathworld.wolfram.com/FourierTransform.html
 

Offline syhprum

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The momentum operator
« Reply #4 on: 08/06/2008 14:24:53 »
My thanks to Bored Chemist and Jpetruccelli for drawing my attention to the two under mentioned sites that gave me an hour of very interesting reading especially the problem of the cattle that required an unthinkable number to solve! and the various bizarre experiments.

http://mathworld.wolfram.com

http://en.wikipedia.org/wiki/MythBusters
 

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The momentum operator
« Reply #4 on: 08/06/2008 14:24:53 »

 

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